NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Edited By Ramraj Saini | Updated on Nov 30, 2023 09:53 AM IST

NCERT Solutions for Class 6 Maths Chapter 8 Decimals are discussed here. Subject matter at Careers360 has created detailed solutions considering the latest syllabus of CBSE 2023. Students can check detailed NCERT Class 6 Maths solutions for chapter 8 in this article. There are many questions in the Class 6 solutions of NCERT Maths chapter 8 which will give more clarity to the concepts. In this NCERT 6 Maths solutions chapter 8 students will study about addition and subtraction of decimal numbers and other concepts.

Latest: Important Maths Formulas for Class 6 - Chapterwise

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Formulae
NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Points
NCERT Solutions for Class 6 Maths Chapter 8 Decimals
NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Intext Questions and Exercise)
Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Subtopic Fractions as Decimals
NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.1
NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.2
CBSE NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.3
Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.5 Using Decimals
Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.2 Length
Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.3 Weight
NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.4
Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.6 Addition of Numbers with Decimals
NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.5
Question:1 Find the sum in each of the following :
Solutions for NCERT Class 7 Maths Chapter 8 Decimals Topic 8.7 Subtraction of Decimals
Question:1 Subtract from
CBSE NCERT Solutions for Class 7 Maths Chapter 8 Decimals-Exercise: 8.6
Decimals Class 6 Maths Chapter 8-Topics
NCERT Solutions for Class 6 Mathematics - Chapter-Wise
NCERT Solutions for Class 6 Mathematics - Subject Wise

$NCERT Solutions for Class 6 Maths Chapter 8 Decimals$

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

There are 38 questions in 6 exercises of this chapter of NCERT. You will get detailed explanations of these questions in the CBSE NCERT solutions for Class 6 Maths chapter 8. Referring to NCERT Class 6 Maths solutions of chapter 8 helps to understand the chapters of NCERT Class 6 Syllabus in a more easy way. Topic-wise and exercise-wise solved questions are available below in the NCERT solutions for Class 6 Maths chapter 8 Decimals. You can also access NCERT Solutions for all Classes.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Formulae

Tenths = 1/10
Hundredths = 1/100
The decimal point is moved to the left by the number of zeros in the denominator

For example 150/100 = 1.5 (shifting decimal point by 2 points to left)

For more, Download Ebook - NCERT Class 6 Maths: Chapterwise Important Formulas And Points

NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Points

Every decimal can be written as a fraction: Decimals can be expressed as fractions, where the decimal part is written as the numerator and the place value of the decimal determines the denominator. For example, 0.5 can be written as 5/10 or 1/2.

Comparing decimal numbers: Decimal numbers can be compared by starting with the whole part. If the whole parts are equal, then the tenth parts can be compared, followed by the hundredth parts, and so on. By comparing the place values, you can determine which decimal number is greater or smaller.

Use of decimals: Decimals are used in various real-life situations, including money transactions, measurements, scores in sports, and scientific calculations. They provide a more precise representation of values compared to whole numbers.

Free download NCERT Solutions for Class 6 Maths Chapter 8 Decimals for CBSE Exam

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

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NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Intext Questions and Exercise)

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Tenths
Question: 1 Can you now write the following as decimals

Hundreds (100)	Tens (10)	Ones (1)	Tenths $\small \left ( \frac{1}{10} \right )$
5	3	8	1
2	7	3	4
3	5	4	6

Answer: Yes, we can write to them in Decimal form. The Numbers in the decimal forms are:

i) 538.1
ii) 273.4
iii) 354.6

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Subtopic Fractions as Decimals

Question: 1 Write $\small \frac{3}{2},\frac{4}{5},\frac{8}{5}$ in decimal notation.

Answer: The number in decimal forms are:

$\frac{3}{2}=1.5$
$\frac{4}{5}=0.8$
$\frac{8}{5}=1.6$

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.1

Question: 1 Write the following as numbers in the given table.

1643089279385 1643089356027

Answer: Observing from the figure, we get

Hundreds

(100)

Tens

(10)

Ones

(1)

Tenths

$\small (\frac{1}{10})$

a) 0

b) 1

Question: 2 Write the following decimals in the place value table.

(a) $\small 19.4$ (b) $\small 0.3$ (c) $\small 10.6$ (d) $\small 205.9$

Answer: The Value table of the numbers are:

Number	Hundreds	Tens	Ones	Tenths
19.3	0	1	9	3
0.3	0	0	0	3
10.6	0	1	0	6
205.9	2	0	5	9

Question: 3 Write each of the following as decimals :

(a) Seven-tenths (b) Two tens and nine-tenths

(e) Six hundred point eight

Answer:

(a) Seven-tenths :
- $7\times\frac{1}{10}=\frac{7}{10}=0.7$
(b) Two tens and nine-tenths:
- $2\times10+9\times\frac{1}{10}=20+\frac{9}{10}=20+0.9=20.9$
(c) Fourteen point six :
- $=14.6$
(d) One hundred and two ones
- $1\times100+2\times1=100+2=102.0$
(e) Six hundred point eight
- $=600.8$

Question: 4 Write each of the following as decimals:

(a) $\small \frac{5}{10}$ (b) $\small 3+\frac{7}{10}$ (c) $\small 200+60+5+\frac{1}{10}$ (d) $\small 70+\frac{8}{10}$
(e) $\small \frac{88}{10}$ (f) $\small 4\frac{2}{10}$ (g) $\small \frac{3}{2}$ (h) $\small \frac{2}{5}$ (i) $\small \frac{12}{5}$ (j) $\small 3\frac{3}{5}$ (k) $\small 4\frac{1}{2}$

Answer: As we know when we divide a number by 10, the result is that number with a decimal after one digit,(we count the digit from right to left). So Keeping that in mind,

(a) $\small \frac{5}{10}=0.5$
(b) $\small 3+\frac{7}{10}=3+0.7=3.7$
(c) $\small 200+60+5+\frac{1}{10}=265+0.1=265.1$
(d) $\small 70+\frac{8}{10}=70+0.8=70.8$
(e) $\small \frac{88}{10}=8.8$
(f) $\small 4\frac{2}{10}=4+\frac{2}{10}=4+0.2=4.2$
(g) $\small \frac{3}{2}=\frac{3}{2}\times\frac{5}{5}=\frac{15}{10}=1.5$
(h) $\small \frac{2}{5}=\frac{2}{5}\times\frac{2}{2}=\frac{4}{10}=0.4$
(i) $\small \frac{12}{5}=\frac{12}{5}\times\frac{2}{2}=\frac{24}{10}=2.4$
(j) $\small 3\frac{3}{5}=3+\frac{3}{5}=3+\frac{3}{5}\times\frac{2}{2}=3+\frac{6}{10}=3+0.6=3.6$
(k) $\small 4\frac{1}{2}=4+\frac{1}{2}=4+\frac{1}{2}\times\frac{5}{5}=4+\frac{5}{10}=4+0.5=4.5$

Question: 5 Write the following decimals as fractions. Reduce the fractions to lowest form.

(a) $\small 0.6$ (b) $\small 2.5$ (c) $\small 1.0$ (d) $\small 3.8$ (e) $\small 13.7$ (f) $\small 21.2$ (g) $\small 6.4$

Answer: Converting Decimals into Fractions. we get

(a) $\small 0.6=\frac{6}{10}=\frac{3}{5}$
(b) $\small 2.5=\frac{25}{10}=\frac{5}{2}$
(c) $\small 1.0=1$
(d) $\small 3.8=\frac{38}{10}=\frac{19}{5}$
(e) $\small 13.7=\frac{137}{10}$
(f) $\small 21.2=\frac{212}{10}=\frac{106}{5}$
(g) $\small 6.4=\frac{64}{10}=\frac{32}{5}$

Question: 6 Express the following as cm using decimals.

(a) 2 mm (b) 30 mm (c) 116 mm (d) 4 cm 2 mm (e) 162 mm (f) 83 mm

Answer: As we know
1 cm = 10 mm
1 mm = 0.1 cm
So,

(a) 2 mm $=\frac{2}{10}\:cm=0.2\:cm$

(b) 30 mm $=\frac{30}{10}\:cm=3.0\: cm$

(d) 4 cm 2 mm $=4\:cm+\frac{2}{10}\:cm=4+0.2=4.2\: cm$

(e) 162 mm $=\frac{162}{10}\:cm=16.2\: cm$

(f) 83 mm $=\frac{83}{10}\:cm=8.3\: cm$

Question: 7 Between which two whole numbers on the number line are the given numbers lie? Which of these whole numbers is nearer the number?

1643089406018

(a) $\small 0.8$ (b) $\small 5.1$ (c) $\small 2.6$ (d) $\small 6.4$ (e) $\small 9.1$ (f) $\small 4.9$

Answer:

(a) $\small 0.8$ Lies Between 0 and 1. and 1 is the closest whole number to it.
(b) $\small 5.1$ Lies between 5 and 6 and 5 is the closest whole number to it.
(c) $\small 2.6$ Lies between 2 and 3 and 3 is the closest whole number to it.
(d) $\small 6.4$ Lies between 6 and 7 and 6 is the closest whole number to it.
(e) $\small 9.1$ Lies between 9 and 10 and 9 is the closest whole number to it.
(f) $\small 4.9$ Lies between 4 and 5. and 5 is the closest whole number to it.

Question: 8 Show the following numbers on the number line.

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

Answer: The numbers on the number line are:

(a) $\small 0.2$ (b) $\small 1.9$ (c) $\small 1.1$ (d) $\small 2.5$

1643089447869

Question: 9 Write the decimal number represented by the points A, B, C, D on the given number line.

1643089479134

Answer: As we can see the points lies in the number line,

$A=\frac{8}{10}=0.8$

$B=1+\frac{3}{10}=1+0.3=1.3$

$C=2+\frac{2}{10}=2+0.2=2.2$

$D=2+\frac{9}{10}=2+0.9=2.9$

Question: 10 (a) The length of Ramesh’s notebook is 9 cm 5 mm. What will be its length in cm?

Answer: length of Ramesh’s notebook = 9 cm 5 mm

As we know,

$1mm=\frac{1}{10}cm$

So Length in the unit of cm :

$=9cm+5\times\frac{1}{10}cm=9cm+0.5cm=9.5cm$

Hence the Length of Ramesh's notebook is 9.5 cm.

Question: 10 (b) The length of a young gram plant is 65 mm. Express its length in cm.

Answer: length of a young gram plant is 65 mm

As we know,

$1mm=\frac{1}{10}cm$

So,

$65mm=65\times\frac{1}{10}cm=\frac{65}{10}cm=6.5cm$

So Length of young gram plant is 6.5 cm.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.2

Question:1 Complete the table with the help of these boxes and use decimals to write the number.

1643089584450

1643089614262

	Ones	Tenths	Hundredsths	Number
(a)
(b)
(c)

Answer:

	Ones	Tenths	Hundredths	Number
(a)	0	2	6	0.26
(b)	1	3	8	1.38
(c)	1	2	8	1.28

Question:2 Write the numbers given in the following place value table in decimal form.

Hundreds
100
Tens
10
Ones
1
Tenths
$\small \frac{1}{10}$
Hundredths
$\small \frac{1}{100}$
Thousands
$\small \frac{1}{1000}$
(a)
(b)
(c)
(d)
(e)
0
1
0
2
0
0
0
3
1
1
3
2
0
1
2
2
6
0
9
2
5
3
2
0
4
0
0
5
2
1
Answer: As we can see from the table, the numbers are:

(a) $3+\frac{2}{10}+\frac{5}{100}=3+0.2+0.05=3.25$

(b) $100+2+\frac{6}{10}+\frac{3}{100}=100+2+0.6+0.03=102.63$

(d) $200+10+1+\frac{9}{10}+\frac{2}{1000}=211+0.9+0.002=211.902$

(e) $10+2+\frac{2}{10}+\frac{4}{100}+\frac{1}{1000}=12+0.2+0.04+0.001=12.241$

Question: 3 Write the following decimals in the place value table.

(a) $\small 0.29$ (b) $\small 2.08$ (c) $\small 19.60$ (d) $\small 148.32$ (e) $\small 200.812$

Answer:

(a)

$\small 0.29=0.2+0.09=\frac{2}{10}+\frac{9}{100}$

(b)

$\small 2.08=2+0.08=2+\frac{8}{100}$

(c)

$\small 19.60=19+0.6=10+9+\frac{6}{10}$

(d)

$\small 148.32=148+0.3+0.02=100+40+8+\frac{3}{10}+\frac{2}{100}$

(e)

$\small 200.812=200+0.8+0.01+0.002=200+\frac{8}{10}+\frac{1}{100}+\frac{2}{1000}$

So, the table becomes:

Hundreds	Tens	Ones	Tenths	Hundredths	Thousands
0	0	0	2	9	0
0	0	2	0	8	0
0	1	9	6	0	0
1	4	8	3	2	0
2	0	0	8	1	2

Question: 4 Write each of the following as decimals.

(a) $\small 20+9+\frac{4}{10}+\frac{1}{100}$ (b) $\small 137+\frac{5}{100}$ (c) $\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}$
(d) $\small 23+\frac{2}{10}+\frac{6}{1000}$ (e) $\small 700+20+5+\frac{9}{100}$

Answer:

Writing the number in a decimal form:

(a)

$\small 20+9+\frac{4}{10}+\frac{1}{100}=29+0.4+0.01=29.41$

(b)

$\small 137+\frac{5}{100}=137+0.05=137.05$

(c)

$\small \frac{7}{10}+\frac{6}{100}+\frac{4}{1000}=0.7+0.06+0.004=0.764$
(d)

$\small 23+\frac{2}{10}+\frac{6}{1000}=23+0.2+0.006=23.206$

(e)

$\small 700+20+5+\frac{9}{100}=725+0.09=725.09$

Question: 5 Write each of the following decimals in words.

(a) $\small 0.03$ (b) $\small 1.20$ (c) $\small 108.56$ (d) $\small 10.07$ (e) $\small 0.032$ (f) $\small 5.008$

Answer: As we know After decimal, we call one digit at a time.

(a) $\small 0.03$ = Zero point zero three
(b) $\small 1.20$ = One point two zero
(c) $\small 108.56$ = One hundred eight point five six
(d) $\small 10.07$ = Ten point zero seven
(e) $\small 0.032$ = Zero point zero three two
(f) $\small 5.008$ = Five point zero zero eight.

Question:6 Between which two numbers in tenths place on the number line does each of the given number lies?

(a) $\small 0.06$ (b) $\small 0.45$ (c) $\small 0.19$ (d) $\small 0.66$ (e) $\small 0.92$ (f) $\small 0.57$

Answer: Here, we have a zoomed version of the number line in which the interval between two number is 0.1 instead of 1 which we use normally. So,

(a) $\small 0.06$ lies between 0.0 and 0.1
(b) $\small 0.45$ lies between 0.4 and 0.5
(c) $\small 0.19$ lies between 0.1 and 0.2
(d) $\small 0.66$ lies between 0.6 and 0.7
(e) $\small 0.92$ lies between 0.9 and 1.0
(f) $\small 0.57$ lies between 0.5 and 0.6.

Question:7 Write as fractions in lowest terms.

(a) $\small 0.60$ (b) $\small 0.05$ (c) $\small 0.75$ (d) $\small 0.18$ (e) $\small 0.25$ (f) $\small 0.125$ (g) $\small 0.066$

Answer: The Numbers in the lowest form of the fraction are:

(a)

$\small 0.60=\frac{60}{100}=\frac{6}{10}=\frac{3}{5}$

(b)

$\small 0.05=\frac{5}{100}=\frac{1}{20}$

(c)

$\small 0.75=\frac{75}{100}=\frac{3}{4}$

(d)

$\small 0.18=\frac{18}{100}=\frac{9}{50}$

(e)

$\small 0.25=\frac{25}{100}=\frac{1}{4}$

(f)

$\small 0.125=\frac{125}{1000}=\frac{5}{40}=\frac{1}{8}$

(g)

$\small 0.066=\frac{66}{1000}=\frac{33}{500}$

CBSE NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.3

Question: 1 Which is greater? (a) $\small 0.3$ or $\small 0.4$ (b) $\small 0.07$ or $\small 0.02$ (c) $\small 3$ or $\small 0.8$ (d) $\small 0.5$ or $\small 0.05$ (e) $\small 1.23$ or $\small 1.2$ (f) $\small 0.099$ or $\small 0.19$

(g) $\small 1.5$ or $\small 1.50$ (h) $\small 1.431$ or $\small 1.490$ (i) $\small 3.3$ or $\small 3.300$ (j) $\small 5.64$ or $\small 5.603$

Answer: We compare Decimals just like the way we compare normal numbers. First, we see the leftmost digit of the numbers and if they are same then we move toward the right digits. So,

(a) $\small 0.3$ or $\small 0.4$

0.4 is greater as while number part (the number before the decimal) is the same.

(b) $\small 0.07$ or $\small 0.02$

0.07 is greater

3 is greater as the whole part of the numbers are 3 and 0.so obviously 3 is greater than 0.

(d) $\small 0.5$ or $\small 0.05$

0.5 is greater

(e) $\small 1.23$ or $\small 1.2$

1.23 is greater

(f) $\small 0.099$ or $\small 0.19$

0.19 is greater

(g) $\small 1.5$ or $\small 1.50$

As we know that after the decimal, we can add as many zeros as we want in the rightmost to the numbers just like we can add zeroes to the left side of any whole number. this we can do because those zeroes do not carry any value so they don't affect the numbers.

So. They both are the same number.

(h) $\small 1.431$ or $\small 1.490$

1.490 is greater

(i) $\small 3.3$ or $\small 3.300$

As we know in decimal the RIGHTMOST zeros don't carry any value.so,

They both are the same number

(j) $\small 5.64$ or $\small 5.603$

5.64 is greater.

Question:2 Make five more examples and find the greater number from them.

Answer: we can take any two number from the number line and compare them. For example,

1) 9.9 > 3.3

2) 6.6 < 9.9

3) 3.6 > 3.3

4) 9.6 < 9.9

5) 6.9 > 3.9

Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.5 Using Decimals

Question:(i) Write 2 rupees 5 paise and 2 rupees 50 paise in decimals.

Answer: As we know that

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

2 rupees 5 paise :

$=2+5\times\frac{1}{100}=2+\frac{5}{100}=2+0.05=2.05\:rupees$

and 2 rupees 50 paise:

$=2+50\times\frac{1}{100}=2+\frac{50}{100}=2+0.5=2.5\:rupees$

Question:(ii) Write 20 rupees 7 paise and 21 rupees 75 paise in decimals?

Answer: As we know that

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

20 rupees 7 paise:

$=20+7\times\frac{1}{100}=20+\frac{7}{100}=20+0.07=20.07\:rupees$

and 21 rupees 75 paise:

$=21+75\times\frac{1}{100}=21+\frac{75}{100}=21+0.75=21.75\:rupees$

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.2 Length

Question:1 Can you write 4 mm in ‘cm’ using decimals?

Answer: As we know,

1 cm = 10 mm

$1 mm = \frac{1}{10}\:cm$

So,

$4 mm = 4\times\frac{1}{10}\:cm=\frac{4}{10}\:cm=0.4cm$ .

Question:2 How will you write 7cm 5 mm in ‘cm’ using decimals?

Answer: As we know,

1 cm = 10 mm

$1mm=\frac{1}{10}cm$

So,

7cm 5 mm :

$=7 + 5\times\frac{1}{10}=7+\frac{5}{10}=7+0.5=7.5cm$ .

Question:3 Can you now write 52 m as ‘km’ using decimals? How will you write 340 m as ‘km’ using decimals? How will you write 2008 m in ‘km’?

Answer: As we know,

1 km = 1000 m

$1 m =\frac{1}{1000}km$

So,

52 m in km :

$52 m =52\times\frac{1}{1000}km=\frac{52}{1000}km=0.052km$
340 m in km :

$340m =340\times\frac{1}{1000}km=\frac{340}{1000}km=0.34km$

2008 m in km :

$2008m =2008\times\frac{1}{1000}km=\frac{2008}{1000}km=2.008km$

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.3 Weight

Question:1 Can you now write 456 g as ‘kg’ using decimals?

Answer: As we know,

1 kg = 1000 g

$1\:g=\frac{1}{1000}\:kg$

So,

456 g in kg:

$456\:g=456\times\frac{1}{1000}\:kg=\frac{456}{1000}\:kg=0.456\:kg$

Question:2 How will you write 2 kg 9 g in ‘kg’ using decimals?

Answer: As we know,

1 kg = 1000 g

$1\:g = \frac{1}{1000}\:kg$

So,

2 kg 9 g in kg:

$2\:kg+9\:g = 2\:kg+\frac{9}{1000}\:kg=2\:kg+0.009\:kg=2.009\:kg$

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.4

Question:1 Express as rupees using decimals.

(a) 5 paise (b) 75 paise (c) 20 paise (d) 50 rupees 90 paise (e) 725 paise

Answer: As we know,

1 rupee = 100 paise

$1\:paise=\frac{1}{100}\:rupees$

(a) 5 paise

$=5\times\frac{1}{100}=\frac{5}{100}=0.05\:rupees$

(b) 75 paise

$=75\times\frac{1}{100}=\frac{75}{100}=0.75\:rupees$

$=20\times\frac{1}{100}=\frac{20}{100}=0.20\:rupees$

(d) 50 rupees 90 paise

$=50+90\times\frac{1}{100}=50+\frac{90}{100}=50+0.9=50.9\:rupees$

(e) 725 paise

$=725\times\frac{1}{100}=\frac{725}{100}=7.25\:rupees$

Question:2 Express as meters using decimals.

(a) 15 cm (b) 6 cm (c) 2 m 45 cm (d) 9 m 7 cm (e) 419 cm

Answer: As we know,

1 m = 100 cm

Which means

$1\:cm=\frac{1}{100}\:m$ .

So,

(a) 15 cm

$15\:cm=15\times\frac{1}{100}=\frac{15}{100}=0.15\:m$

(b) 6 cm

$6\:cm=6\times\frac{1}{100}=\frac{6}{100}=0.06\:m$

$2+45\:cm=2+45\times\frac{1}{100}=2+\frac{45}{100}=2+0.45=2.45\:m$

(d) 9 m 7 cm

$9+7\:cm=9+7\times\frac{1}{100}=9+\frac{7}{100}=9+0.07=9.07\:m$

(e) 419 cm

$419\:cm=419\times\frac{1}{100}=\frac{419}{100}=4.19\:m$

Question:3 Express as cm using decimals.

(a) 5 mm (b) 60 mm (c) 164 mm (d) 9 cm 8 mm (e) 93 mm

Answer: As we know,

1 cm = 10 mm

which means

$1\:mm=\frac{1}{10}\:cm$

So,

(a) 5 mm :

$5\:mm=5\times\frac{1}{10}=\frac{5}{10}=0.5\:cm$

(b) 60 mm

$60\:mm=60\times\frac{1}{10}=\frac{60}{10}=6.0\:cm$

$164\:mm=164\times\frac{1}{10}=\frac{164}{10}=16.4\:cm$

(d) 9 cm 8 mm

$9\:cm+8\:mm=9+8\times\frac{1}{10}=9+\frac{8}{10}=9+0.8=9.8\:cm$

(e) 93 mm

$93\:mm=93\times\frac{1}{10}=\frac{93}{10}=9.3\:cm$

Question:4 Express as km using decimals.

(a) 8 m (b) 88 m (c) 8888 m (d) 70 km 5 m

Answer: As we know,

1 km = 1000 m

which means

$1\:m=\frac{1}{1000}\:km$

So,

(a) 8 m

$8\:m=8\times\frac{1}{1000}=\frac{8}{1000}=0.008\:km$

(b) 88 m

$88\:m=88\times\frac{1}{1000}=\frac{88}{1000}=0.088\:km$

$8888\:m=8888\times\frac{1}{1000}=\frac{8888}{1000}=8.888\:km$

(d) 70 km 5 m

$70\:km+5\:m=70+5\times\frac{1}{1000}=70+\frac{5}{1000}=70+0.005=70.005\:km$

Question:5 Express as kg using decimals.

(a) 2 g (b) 100 g (c) 3750 g
(d) 5 kg 8 g (e) 26 kg 50 g

Answer: As we know,

1 kg = 1000 g

which means

$1 g = \frac{1}{1000}kg$

So,

(a) 2 g

$2 g = \frac{2}{1000}kg=0.002kg$

(b) 100 g

$100 g = \frac{100}{1000}kg=0.1kg$

$3750 g = \frac{3750}{1000}kg=3.75kg$

(d) 5 kg 8 g

$5kg\:\:8g=5+8g =5kg+ \frac{8}{1000}kg=5kg+0.008kg=5.008kg$

(e) 26 kg 50 g

$26kg\:\:50g=26+50g =26kg+ \frac{50}{1000}kg=26kg+0.05kg=26.05kg$

Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.6 Addition of Numbers with Decimals

Question:(i) Find

$\small 0.29 + 0.36$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$\small \\0.29$

$\small +$ $\small \\0.36$

$\small \\0.65$

Hence

$\small 0.29 + 0.36=0.65$

Question:(ii) Find

$\small 0.7 + 0.08$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$0.70$

$+\:0.08$

$0.78$

Hence,

$\small 0.7 + 0.08=0.78$

Question:(iii) Find

$\small 1.54 + 1.80$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$1.54$

$+\:1.80$

$3.34$ .

Question:(iv) Find

$\small 2.66 + 1.85$

Answer: As we know that decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

$2.66$

$+\:1.85$

$4.51$

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.5

Question:1 Find the sum in each of the following :

(a) $\small 0.007 + 8.5 + 30.08$

(b) $\small 15 + 0.632 + 13.8$

(d) $\small 25.65 + 9.005 + 3.7$

(e) $\small 0.75 + 10.425 + 2$

(f) $\small 280.69 + 25.2 + 38$

Answer: As we know that the decimal addition is exactly like normal addition and we don't change the position of decimal in the number.

So,

(a) $\small 0.007 + 8.5 + 30.08$

$0.007$

$8.500$

$+\:30.080$

$38.587$

(b) $\small 15 + 0.632 + 13.8$

$\small 15.000$

$\small 0.632$

$\small +\:13.800$

$\small 29.432$

$27.076$

$0.550$

$+\:0.004$

$27.630$

(d) $\small 25.65 + 9.005 + 3.7$

$25.650$

$9.005$

$+\:3.700$

$38.355$

(e) $\small 0.75 + 10.425 + 2$

$0.750$

$10.425$

$+\:2.000$

$13.175$

(f) $\small 280.69 + 25.2 + 38$

$280.69$

$25.20$

$+\:38.00$

$343.89$

Question: 2 Rashid spent Rs $\small 35.75$ for Maths book and Rs $\small 32.60$ for Science book. Find the total amount spent by Rashid.

Answer:

Amount spend on maths book = Rs $\small 35.75$

Amount spend on science book = Rs $\small 32.60$

Total amount spend = Rs $\small 35.75$ $+$ Rs $\small 32.60$

= Rs $68.35$

Hence the total amount spends by Rashid is Rs 68.35.

Question:3 Radhika’s mother gave her Rs $\small 10.50$ and her father gave her Rs $\small 15.80$ , find the total amount given to Radhika by the parents.

Answer:

The amount Radhika got from mother = Rs $\small 10.50$

The amount Radhika got from father = Rs $\small 15.80$

Total amount Radhika have = Rs $\small 10.50$ $+$ Rs $\small 15.80$

= Rs $26.30$

Hence the total amount of money Radhika got from her parents is Rs 26.30.

Question:4 Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.

Answer:

Length of cloth for shirt = 3 m 20 cm

Length of cloth for trouser = 2 m 5 cm

As we know.

For adding two numbers their unit must be the same.

So making length in meter:

$3m+20cm=3\:m+\frac{20}{100}m$

$=3\:m+0.2m$

$=3.2m$

and

$2m+5cm=2\:m+\frac{5}{100}m$

$=2\:m+0.05m$

$=2.05m$

Now,

Total Length of the cloth = 3.2 m + 2.05 m

= 5.25 m

Hence Nasreen bought clothe with a length of 5.25 m.

Question:5 Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?

Answer:

The distance covered in the morning = 2 km 35 m

The distance covered in the evening = 1 km 7 m

As we know.

For adding two numbers their unit must be the same.

So Lengths in the Kilometer :

$2 km+35m=2km+\frac{35}{1000}km$

$=2km+0.035km$

$=2.035km$

And

$1km+7m=1km+\frac{7}{1000}km$

$=1km+0.007km$

$=1.007\:km$

Now,

Total Distance covered by Naresh = 2.035 km + 1.007 km

= 3.042 km

Hence the distance covered by Naresh is 3.042 km.

Question:6 Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?

Answer:

The distance travelled by bus = 15 km 268 m

The distance travelled by car = 7 km 7 m

The distance travelled by foot = 500 m

As we know.

For adding two numbers their unit must be the same.

So,

$15 km +268 m = 15 km + \frac{268}{1000}km$

$= 15 km + 0.268km$

$= 15.268km$

Also

$7km+7m=7km+\frac{7}{1000}km$

$=7km+0.007km$

$=7.007km$

And,

$500m=\frac{500}{1000}km$

$=0.5km$

So,

The total distance Sunita traveled = 15.268 km + 7.007 km + 0.5 km

= 22.775 km

Hence the distance between Sunita's residence and school is 22.775 km.

Question:7 Ravi purchased 5 kg 400 g rice, 2 kg 20 g sugar and 10 kg 850g flour. Find the total weight of his purchases.

Answer:

amount of rice purchased by Ravi = 5 kg 400 g.

amount of sugar purchased by Ravi = 2 kg 20 g.

amount of flour purchased by Ravi = 10 kg 850 g.

As we know.

For adding two numbers their unit must be the same.

So,

$5\:kg+400g=5kg+\frac{400}{1000}kg$

$=5kg+0.4kg$

$=5.4kg$

Also,

$2\:kg+20g=2kg+\frac{20}{1000}kg$

$=2kg+0.02kg$

$=2.002kg$

And,

$10\:kg+850g=10kg+\frac{850}{1000}kg$

$=10kg+0.85kg$

$=10.85kg$

So,

The total weight of the grain Ravi purchased = 5.4 kg + 2.002 kg + 10.85 kg

= 18.270 kg

Hence Ravi purchased total grain having the weight of 18.270 kg.

Solutions for NCERT Class 7 Maths Chapter 8 Decimals Topic 8.7 Subtraction of Decimals

Question:1 Subtract $\small 1.85$ from $\small 5.46$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

$5.46$

$-1.85$

$3.61$

Question:2 Subtract $\small 5.25$ from $\small 8.28$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

$8.28$

$-5.25$

$3.03$

Question:3 Subtract $\small 0.95$ from $\small 2.29$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

$2.29$

$-0.95$

$1.34$

Question: 4 Subtract $\small 2.25$ from

$\small 5.68$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

$5.68$

$-2.25$

$3.43$

CBSE NCERT Solutions for Class 7 Maths Chapter 8 Decimals-Exercise: 8.6

Question: 1 Subtract :

(a) Rs $\small 18.25$ from Rs $\small 20.75$

(b) $\small 202.54$ m from 250 m

(d) $\small 2.051$ km from $\small 5.206$ km

(e) $\small 0.314$ kg from $\small 2.107$ kg

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

(a) Rs $\small 18.25$ from Rs $\small 20.75$

$20.75$

$-18.25$

$2.50$ Rs

(b) $\small 202.54$ m from 250 m\

$250.00$

$-202.54$

$47.46$ m

(c)Rs $\small 5.36$ from Rs $\small 8.40$

$8.40$

$-5.36$

$3.04$ Rs

(d ) $\small 2.051$ km from $\small 5.206$ km

$5.206$

$-2.051$

$3.155$ km

(e) $\small 0.314$ kg from $\small 2.107$ kg

$2.107$

$-0.314$

$1.793$ kg

Question: 2 Find the value of :

(a) $\small 9.756 - 6.28$

(b) $\small 21.05 -15.27$

(d) $\small 11.6 - 9.847$

Answer: As we know, subtraction in decimal works exactly like normal subtraction, we just have to maintain the place of decimal.

(a) $\small 9.756 - 6.28$

$9.756$

$-6.280$

$3.476$

(b) $\small 21.05 -15.27$

$21.05$

$-15.27$

$5.78$

$18.50$

$-6.79$

$11.71$

(d) $\small 11.6 - 9.847$

$11.600$

$-9.847$

$1.753$

Question: 3 Raju bought a book for Rs $\small 35.65$ . He gave Rs 50 to the shopkeeper. How much money did he get back from the shopkeeper?

Answer: Total money Raju gave to shopkeeper = Rs 50

The price of the book = Rs 35.65

Money shopkeeper will return Raju = Rs 50.00 - Rs 35.65

= Rs 14.35

Hence Raju got 14.35 Rs from the shopkeeper.

Question: 4 Rani had Rs $\small 18.50$ . She bought one ice-cream for Rs $\small 11.75$ . How much money does she have now?

Answer: Total Money Rani had = Rs 18.50

The money she spent on ice-cream = 11.75

The Remaining money Rani have = Rs 18.50 - Rs 11.75

= Rs 6.75

Hence Rani has 6.75 Rs now.

Question:5 Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?

Answer: Length of cloth initially = 20 m 5 cm = 20.05 m

Length of cloth Tina cuts = 4 m 50 cm = 4.50 m

The length of remaining cloth = 20.05 m - 4.50 m

= 15.55 m

= 15 m 55 cm

Hence the length of remaining cloth Tina have is 15 m 55 cm.

Question:6 Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?

Answer: Total distance travelled by Namita = 20 km 50 m = 20.05 km

The distance travelled by bus = 10 km 200 m = 10.20 km

The remaining distance which is traveled by auto = 20.05 km - 10.20 km

= 9.850 km.

Hence the distance travelled by Namita in auto is 9.850 km.

Question:7 Aakash bought vegetables weighing 10 kg. Out of this, 3 kg 500 g is onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?

Answer: Total weight of the vegetables = 10 kg

weight of onion = 3 kg 500 g = 3.5 kg

weight of tomato = 2 kg 75 g = 2.075 kg

Total weight of onion and potato = 3.5 kg + 2.075 kg

= 5.575 kg

So, The rest weight of potato = 10 kg - 5.575 kg

= 4.425 kg

Hence, the weight of the potato is 4.425 kg.

Decimals Class 6 Maths Chapter 8-Topics

Tenths
Hundredths
Comparing Decimals
Using Decimals
Addition of Numbers with Decimals
Subtraction of Decimals

NCERT Solutions for Class 6 Mathematics - Chapter-Wise

Chapters No.	Chapters Name
Chapter - 1	Knowing Our Numbers
Chapter - 2	Whole Numbers
Chapter - 3	Playing with Numbers
Chapter - 4	Basic Geometrical Ideas
Chapter - 5	Understanding Elementary Shapes
Chapter - 6	Integers
Chapter - 7	Fractions
Chapter - 8	Decimals
Chapter - 9	Data Handling
Chapter -10	Mensuration
Chapter -11	Algebra
Chapter -12	Ratio and Proportion
Chapter -13	Symmetry
Chapter -14	Practical Geometry

NCERT Solutions for Class 6 Mathematics - Subject Wise

NCERT Solutions for Class 6 Maths

Solutions of NCERT for Class 6 Science

Some important findings from the NCERT solutions for Class 6 Maths chapter 8 Decimals which will be useful in real life too are listed below.

Converting a quantity from one unit to another-

Millimeter to centimeter-

$1mm=\frac{1}{10}cm=0.1cm$

Centimeter to meter-

$1cm=\frac{1}{100}m=0.01m$

Paise to rupee

$1\ paise=\frac{1}{100}rupees=0.01 rupees$ .

$250\ paise=\frac{250}{100}rupees=2.5rupees$

Grams to kg

$1gm=\frac{1}{1000}kg=0.001kg$

Tip- You don't have to remember these units conversion. If you know the value of the quantity in one unit and other units, you can convert it very easily. In CBSE NCERT solutions for class 6 maths chapter 8 decimals, you will find many unit conversion questions that will further strengthen your fundamentals.

Happy Reading !!!

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NCERT Solutions for Class 6 Science Chapter 2 Components of Food

NCERT Solutions for Class 6 Science Chapter 1 Food Where Does It Come From

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Formulae

NCERT Solutions for Class 6 Maths Chapter 8 Decimals - Important Points

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals (Intext Questions and Exercise)

NCERT Solutions for Class 6 Maths Chapter 8 Decimals

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Tenths Question: 1 Can you now write the following as decimals

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Subtopic Fractions as Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.1

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.2

Hundreds100Tens10Ones1TenthsHundredthsThousands(a)(b)(c)(d)(e)010200031132012260925320400521 Answer: As we can see from the table, the numbers are:

CBSE NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.3

Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.5 Using Decimals

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.2 Length

Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.5.3 Weight

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.4

Solutions for NCERT Class 6 Maths Chapter 8 Decimals Topic 8.6 Addition of Numbers with Decimals

NCERT Solutions for Class 6 Maths Chapter 8 Decimals-Exercise: 8.5

Question:1 Find the sum in each of the following :

Solutions for NCERT Class 7 Maths Chapter 8 Decimals Topic 8.7 Subtraction of Decimals

Question:1 Subtract from

CBSE NCERT Solutions for Class 7 Maths Chapter 8 Decimals-Exercise: 8.6

Decimals Class 6 Maths Chapter 8-Topics

NCERT Solutions for Class 6 Mathematics - Chapter-Wise

NCERT Solutions for Class 6 Mathematics - Subject Wise

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Solutions of NCERT for Class 6 Maths Chapter 8 Decimals Topic 8.2 Tenths
Question: 1 Can you now write the following as decimals

Hundreds
100
Tens
10
Ones
1
Tenths
$\small \frac{1}{10}$
Hundredths
$\small \frac{1}{100}$
Thousands
$\small \frac{1}{1000}$
(a)
(b)
(c)
(d)
(e)
0
1
0
2
0
0
0
3
1
1
3
2
0
1
2
2
6
0
9
2
5
3
2
0
4
0
0
5
2
1
Answer: As we can see from the table, the numbers are:

Question:1 Subtract $\small 1.85$ from $\small 5.46$